Useful Definition: $f$ is continuous at $x_o\in X$ if for every $\epsilon >0, \exists \delta = \delta(\epsilon,x_o)<0$ such that if $x \in X$ and $\|x-x_o\|<\delta$ then $\|f(x)-f(x_o)\|<\epsilon$
$f$ is continuous if it is continuous for all $x\in X$.
Useful Theorem: $f$ is continuous (on all of $x$) $\iff$ for every open set $V$ in $R^m$, $f^{-1}(V)$ is open in $X$.
Exercise: Let $f:X\rightarrow Y$. Show that $f$ is continuous if and only if for each $x \in X$ there is a neighborhood $U$ of $x$ such that $f|U$ is continuous.
My attempt:
proof.
($\implies$) Suppose that $f$ is continuous $\forall x\in X$. Then there exists some open set in $V\subset Y$ such that $U \subset f^{-1}(V)$ where $f^{-1}(V)$ is open. Now since $U$ is a neighborhood it is open (I've proved that neighborhoods are open previously). So by the Theorem above $f|U$ is continuous.
($\Longleftarrow$) Suppose that for each $x \in X$ there is a neighborhood $U$ of $x$ such that $f|U$ is continuous…
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I'm not quite sure how to prove the rest of this or whether my first part of the proof was correct. Any guidance would be appreciated.
Best Answer
I guess this exercise is the exercise 5 in section 3 in chapter 1 on p.30 in "Analysis on Manifolds" by James R. Munkres.
On p.26 in this book, there is the definition of "neighborhood":
My solution is here: